How do you solve Number of Visible People in a Queue on LeetCode?

Number of Visible People in a Queue (LeetCode #1944) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a stack helps manage visibility efficiently.

What is the time complexity of Number of Visible People in a Queue?

The optimal solution for Number of Visible People in a Queue runs in O(n) time and uses O(n) space. The time complexity is linear because each person is pushed and popped from the stack at most once.

What is the brute force solution for Number of Visible People in a Queue?

The brute force approach for Number of Visible People in a Queue is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we check for each person how many people they can see to their right by looking at each person in between. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Visible People in a Queue?

Number of Visible People in a Queue uses the following concepts: Array, Stack, Monotonic Stack. The recommended patterns to study are: Monotonic Stack, Array.

What are the interview tips for Number of Visible People in a Queue?

Explain your thought process clearly while coding. Consider edge cases and how they affect your solution. Practice writing clean and efficient code before the interview.

What are common mistakes when solving Number of Visible People in a Queue?

Not handling edge cases like the last person in the queue. Confusing the conditions for visibility.

#1944

Number of Visible People in a Queue

Hard

ArrayStackMonotonic StackMonotonic StackArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Using a monotonic stack allows us to efficiently track the heights of people we can see, reducing the need for nested loops.

⚙️

Algorithm

4 steps

1Step 1: Initialize an answer array of the same length as heights and an empty stack.
2Step 2: Iterate through the heights from right to left.
3Step 3: For each height, pop from the stack until the top of the stack is shorter than the current height, counting how many people can be seen.
4Step 4: Push the current height onto the stack.

solution.py10 lines

1def canSeePeople(heights):
2    n = len(heights)
3    answer = [0] * n
4    stack = []
5    for i in range(n - 1, -1, -1):
6        while stack and heights[stack[-1]] < heights[i]:
7            answer[i] += 1
8            stack.pop()
9        stack.append(i)
10    return answer

ℹ

Complexity note: The time complexity is linear because each person is pushed and popped from the stack at most once.

1Using a stack helps manage visibility efficiently.
2The problem can be visualized as a series of barriers where only taller people can see over shorter ones.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.